# Thread: Problem Sequence

1. ## Problem Sequence

Be a real number that satisfying: $a > 1$. Consider the sequence defined by:

$X_1 = 1$; $X_{n+1} = [\frac{X_n}{2} + \frac{a}{2x_n}]$

1) Prove that $X_2 > X_1$

2. As $X_{n+1} = [\frac{X_n}{2} + \frac{a}{2X_n}]$

Then $X_{2} = [\frac{X_1}{2} + \frac{a}{2X_1}]$

$X_1 = 1$ so

$X_{2} = [\frac{1}{2} + \frac{a}{2\times 1}]$

$X_{2} = [\frac{1+a}{2}]$

If $a=1$ then $X_{2} = \frac{1+1}{2}=1$

as $a>1$ this implies what?

3. Originally Posted by pickslides
As $X_{n+1} = [\frac{X_n}{2} + \frac{a}{2X_n}]$

Then $X_{2} = [\frac{X_1}{2} + \frac{a}{2X_1}]$

$X_1 = 1$ so

$X_{2} = [\frac{1}{2} + \frac{a}{2\times 1}]$

$X_{2} = [\frac{1+a}{2}]$

If $a=1$ then $X_{2} = \frac{1+1}{2}=1$

as $a>1$ this implies what?
Thank you.

2) Show that if $n \ge 2$, $X_n > X_{n-1}$ then $X_{n+1} > X_n$

It is?
$X_1 = 1 ==> n = 1$
$X_2 = 2 ==> n = 2$
$X_n = n$
$X_{n-1} = n-1$
$X_{n+1} = n + 1$
?

4. This is aking for every value after $X_2$ that the sequence is increasing.

Can you show that? You could also use mathematical induction.

5. Originally Posted by pickslides
This is aking for every value after $X_2$ that the sequence is increasing.

Can you show that? You could also use mathematical induction.
Thanks

3) Prove that $X_1 < a$

I can use my solution in problem 1 ?
$X_2 = \frac{1+a}{2}$
$a = 2X_2-1$

$X_2>X_1$ --> $X_1=1 ==> X_2>1$
if $X_2=1 ==> a=1$ then $X_2>1 ==> a>1$
$a>X_1$
Correct ?

6. It is correct ?

7. Originally Posted by Apprentice123
Thanks

3) Prove that $X_1 < a$

I can use my solution in problem 1 ?

I don't think so. In question 1 $X_1 = a$ and that was given so how can it now be less than 1?

8. Originally Posted by pickslides
I don't think so. In question 1 $X_1 = a$ and that was given so how can it now be less than 1?
You're right.
How could I prove ?